1 By shining light through samples, we can learn about the molecular structure of the sample and (sometimes) deduce its structure. IR UV RADIO By the end of this course you will be able to interpret these light absorbing properties of molecules and deduce molecular structures! HC H H H HC H H C H H H HC H H C H H C H H H HC H H C H H C H H C H H H HC H H C C H C H H H H H H C 1 H 4 C 2 H 6 C 3 H 8 C 4 H 10 C 4 H 10 Saturated alkanes (CH only compounds) have the general formula: C N H N+2 Note: it doesn’t matter how the atoms are configured:
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Transcript
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By shining light through samples,we can learn about the molecularstructure of the sample and(sometimes) deduce its structure.
IR
UV
RADIO
By the end of this course you willbe able to interpret these lightabsorbing properties of moleculesand deduce molecular structures!
H CH
HH
H CH
HCHH
H
H CH
HCH
HCH
HH
H CH
HCH
HCH
HCH
HH H C
H
HC
C
HCH
HH
HH
H
C1H4
C2H6
C3H8
C4H10C4H10
Saturated alkanes (CH onlycompounds) have the generalformula:
CNHN+2
Note: it doesn’t matter how theatoms are configured:
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H CH
HCH
HH
H CH
HCH
HCH
HH
C2H6
C3H8
CH
HCH
HCH
HHH C
H
HCH
HCH
HCH
H
C7H16
CNHN+2
C CC
H H
HH
HH
C CH
HH
H
C2H4
C3H6
CH
HCH
HCH
HHH C
H
HCH
HCH
CH
C7H14
CNHN
C CC
H H
HH
HC CH
C2H2
C3H6
C7H12
CCCCCC
H H HH
HH
HHCH
HH
HH
CNHN-2
Two less hydrogens per double bond or ring!
The molecular formula can tell you about the structure:FORMULA TELLS US THE NUMBER OF DOUBLE BONDS ORRINGS:
IHD = index of hydrogen deficiency.a.k.a. Degree of unsaturation.
For CNHM
IHD = 2N+2-M 2
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Example: C6H10
IHD = (6 x 2)+2-10 = 4 = 22 2
•Two double bonds
•Two rings
•One triple bonds
•One double bond plus one ring
C H
F
Cl
Br
I
O N
For each Carbon(valence =4) you add toa structure requires theaddition of two moreHydrogens
Each halogen(valance = 1),takes the place ofa hydrogen
Oxygen(valance = 2)can be addedinto a structurewithoutchanging thenumber ofhydrogensH C
H
HCHH
H
H CH
HCH
HCH
HH
C
Each addednitrogen(valence =3)requires oneadditionalhydrogenX =
4
For CNHXXONIHD = 2N+2-X-(#halogens)+(#Ns) 2
C3H5BrO
Example: IHD =
C4H7NO
Example: IHD =
We can tell something about the molecules you have by thefrequencies of light it absorbs
CH3OH
C C H
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Different frequencies of light (electromagnetic radiation) interact withdifferent aspects molecular motion (potential and kinetic energy).
X-ray:core electronexcitation
UV:valanceelectronicexcitation
IR:molecularvibrations
Radio waves:Nuclear spin states(in a magnetic field)
General Principle of Spectroscopy: A molecule absorbs lightwhen the frequency of light correlates with an energy transition.
Molecules have quantized energy levels:ex. electronic energy levels.
ener
gy
ener
gy
hv
ener
gy } ΔΕ = hv
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Some (not all) nuclei have spin.
Nuclear spin can be examined by Nuclear Magnetic Resonance (NMR)
Only atoms with an odd number of protons or neutrons (or both) havenuclear spins that can be observed by NMR.
=
=En
ergy
In the absence of a magneticfield, both spin states haveequal energy
Ener
gy
No Field Magnetic Field StrongerMagnetic Field
hv
Ener
gy } ΔΕ = hvIn a strong magnetic field, the energylevel difference corresponds to theenergy of radio waves
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How an NMR spectrometerworks:
+ -
N S
RFtransmitter
RFReceiver
Note modernNMRs usesuperconductingmagnets to attainvery strongmagnetic fields
13C-NMRWe can examine the nuclear magnetic properties of carbon atomsin a molecule to learn about a molecules structure.
NOTE: most carbons are 12C. 12C has an even number of protonsand neutrons and cannot be observed by NMR techniques.However approximately 1% of carbons are 13C, and these we cansee in the NMR. (this will be come later when we considernuclear spin coupling)
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A 13C-NMR spectrum.
RF a
bsor
banc
e
RF Frequency
Here is what’s cool about 13C-NMR:
Different carbons appear at differentfrequencies! - Now you know howmany different types of Carbons youhave!
The intensity of the peak doesn’t does not necessarilycorrelate to the number of carbons.
An internalstandardcalled TMSis used toset thescale tozero
Example: Two alcohols (-OH compound) with formula C3H8O1. IHD = 02. Possible structures:
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F
OH
Cl
FF F
How Many 13C NMR resonances?
equivalent resonance structures
=1
BrBr
Br
Br
Br
Br
OH
Cl Br
Br
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An isolated 13C atoms have the same chemical shift. (resonate atthe same frequency)
CNucleus
Electron “cloud”
The electron cloud of the atom partially shields the nucleus from thesurrounding magnetic field
When the carbon nucleus is adjacent to an electronegative atom, thecarbon nuclei has fewer Electrons around it.
C F
Nuclei with less electron density around them will resonate at a differentfrequency: It requires shorter wavelength radio waves (or a weaker magneticfield) to cause a carbon nuclei to resonate if it has less electron density around it.
Conversely atoms that donate electron density cause nuclei to resonate atlower frequencies (higher field strengths).
Si
CH3
CH3
CH3CH3
The silicon in tetramethylsilane (TMS)shields the carbon nuclei and makes themappear “up-field”
The chemical shift spectrum ismeasured relative to TMS.
TMSShift
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TMSShift
δ= is used to signify chemical shift.Because the frequency depends on thefield strength,one uses a relative units scale of “partsper million” or ppm
Chemical shifts reported as ppm units give the same valuesfor the same compound regardless of the instrument used!
δ = (v(compound)- v(TMS))/ v(instrument field strength)
THE δ SCALE IS USED TO MEASURE CHEMICAL SHIFT
HCHHF
HCHHCl
HCHHBr
HCHHOH
HCHHCH3
7.3 ppm
50.2 ppm
71.6 ppm
25.6 ppm
9.6 ppm
HCHHCl
ClCHHCl
ClCClHCl
ClCClClCl
25.6 54.0 77.2 96.1
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H3CCH
CH2
H2C
CH2
CH3
OH
F
H2C
CH2
H2C
CH2
H2C
CH2
H2C
CH3
64
40 27
25
23
2284
30.6
25.3
29.3
29.3
31.9
22.7
14.1
Atoms which are not directly attached to an electronegativeatom are still shifted but the effect is significantly attenuated.The inductive effect (through bond influence) drops off rapidlywith distance.
C O C C C C C
050100150200
40 -5
sp3
90 70sp150 105
sp2
150 120
aromatic
210 160
carbonyl
Typical Values of Chemical shift depend on the1. Hybridization2. Electronegativity of attached atom(s).
C O
C Br, (Cl)
75 50
Some specific values are included in tables 12.1-12.9 in your text book!
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C4H6
14
C3H6O
OH
13C peaks are in reality split by bonded protons.
Generallyshown like this:
Actually lookslike this:
HC
Why??
C H Adjacent nuclear spin of the proton locally increasesmagnetic field felt by carbon nuclei.
Carbons next toup proton spin
Carbons next todown proton spin
NOTE: There are onlytwo possible peaks forC’s having one adjacentproton. or
A “doublet”
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HC Appears as a doublet
HCH
Appears as a triplet
or
There are four possibilities which lead to two net spin states:
The distribution of spin states is random so half the time the carbon will feel no netspin, 25% two up spins and 25% of the time two down spins:
HCHH
Appears as quartet.
There are eight possibilities:
The general rule is:
The number of peaks observed is equal to the number of attached protons,(N), plus one.
Splitting = N+1
FOR 13C it is the number of protons directlyattached to the carbon that cause splitting.
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Why don’t the spins on adjacentCarbons split each other?
C COnly 1% of Carbons have a spin! (Most are 12C and are NMRinactive). Therefore only 1 in ten thousand pairs of Cs have bothhave spin and are coupled. That’s too small to detect!
Note all of these split peaks can quickly get confusing:
Am I looking at two different carbon resonances or a singledoublet?
For this reason we show our spectra without splitting. But we oftensee the splitting indicated over the peak:
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C6H10OIHD = ???
C3H9N44.6 (t)
27.4 (t)
11.5 (q)
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It is helpful to know what typical values of unfunctinalized C’s
Approximate Chemical Shifts ranges for unsubstituted Alkanes-CH3 5-22 ppm-CH2- 15-33 ppm>CH- 25-35
C=C 110-150
Alkyne 66-90
If your molecule contains Oxygen:
OCR H
OCR R
OCR O
OCR NHR
R = alkylOC
R X
X = Cl, Br, I
195 - 220doublet
195 - 220but singlet!
165-180 165-180 165-180
OC
H3C O
OC
H3C NH
60.4 34.4
OC
O O
~155
C OH
C O C
50 - 75 ppm 50 - 75ppm (but two of them!)
CO O
90 - 100 ppm
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If your molecule contains Nitrogen:
C NH2
28-47 ppm
CN
C
H
CN
C
CAmines:
C NO
O61 - 85 ppm
Nitro-
CN RImine
~155 ppm
C NC
14-30
~120
C3H7N139 (d)
113 (t)
44.8 (t)
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C3H9N43.0 (d)
26.5 (q)
C3H8O2
72.7 (d)
67.7 (t)
18.7 (q)
OHHO
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C6H8O
IHDSymmetryWhat is the O doing?What other groups can you identify?Can you account for the IHD?
C8H8OIHDSymmetryWhat is the O doing?What other groups can you identify?
What You Should NowKnow!
• Bond Line Formulas• How to calculate IHDs• 13C chemical shift scale• Multiplicities